Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
Hermite integrators and Riordan arrays
When one feels tired in round-off error
Keigo Nitadori
keigo@riken.jp
Co-Design Team...
Today I talk about
A general form of the correctors for the family of 2-step Hermite
integrators
Up to p-th order derivati...
References
I appreciate for many useful online documents:
1. R. Sprugnoli (2006). An Introduction to Mathematical Methods
...
Polynomial shift by Pascal matrix I
We consider to shift a finite order polynomial f (t), of which up to the
p-th derivativ...
Polynomial shift by Pascal matrix II
A formal solution at t + h is,
F(t + h) = exp

0 1 0 · · · 0
0 0 2 · ...
Polynomial shift by Pascal matrix III
Example
For p = 9,
1 1 1 1 1 1 1 1 1 1
0 1 2 3 4 5 6 7 8 9
0 0 1 3 6 10 15 21 28 36
...
Construction of the Hermite integrators I
For simplicity, we integrate from t = −h to t = h, thus a timestep
becomes ∆t = ...
Construction of the Hermite integrators II
A linear equation we want to solve is
F+
0
F−
1
F+
2
F−
3
...
=
0
0
2
0
4
0
6
0...
Construction of the Hermite integrators III
Example
By solving
F+
0
F−
1
F+
2
=
1 1 1
0 2 4
0 1 6
F0
F2
F4
, (9)
we have
F...
Coefficients table
F+
0 F−
1 F+
2 F−
3 F+
4 F−
5 F+
6 F−
7
A2 1
H4 1 −
1
3
H6 1 −
2
5
2
15
H8 1 −
3
7
4
21
−
2
35
H10 1 −
4
...
General form
For the 2(p + 1)-th order integrator (p ≥ 0), the k-th term
(0 ≤ k ≤ p) is,
c(p)
k =
1
(−2)k
(2k)!!
(2k + 1)!...
Coffee break (double factorial)
Definition:
n!! = n · (n − 2)!! (for n ≥ 2), 1!! = 0!! = 1. (14)
(There are products of odd ...
Outline of the proof
1. We wrote an expected form of the coefficients, c(p)
k
2. Then calculate a differential recurrence, c(...
Differential recurrence I
This is a hand exercise (remember m = (k + 1)/2 ):
c(p)
k − c(p−1)
k =c(p)
k 1 −
p − k + m
p − k
...
Differential recurrence II
Let us now introduce b = k mod 2 ∈ {0, 1}, and ¯b = 1 − b, hence
k = 2m − b. Some useful propert...
Differential recurrence III
Finally, we have
c(p)
k =



1
(−2)p
(2p)!!
(2p + 1)!!
(p = k)
c(p−1)
k +
(−1)k
...
Differential recurrence (old version) I
When k is even (k = 2m),
(2m)!
(2p + 1)!!
p − m − 1
m − 1
(2p − 2m − 1)!!
(2m − 1)!...
Differential recurrence (old version) II
When k is odd (k = 2m − 1),
−
(2m − 1)!
(2p + 1)!!
p − m
m − 1
(2p − 2m − 1)!!
(2m...
Linear equation to solve
What we want to solve is
Ax = b, (A ∈ N
(p+1)×(p+1)
0 , x, b ∈ Qp+1
) (26)
where
Ai j =
2i
j
and ...
LU decomposition
An LU decomposition A = LU is available in,
Ai j =
2i
j
, Li j =
i
j
, Ui j = 22i−j i
j − i
, (29)
irresp...
Formal power series, and coefficient extraction
Formal power series
f (t) =
∞
k=0
tk
fk
f (t) is referred to as a generating...
Proof for A = LU
2i
j
=[tj
](1 + t)2i
=[tj
] 1 + t(t + 2)
i
=[tj
]
∞
k=0
i
k
tk
(t + 2)k
=[tj−k
]
∞
k=0
i
k
∞
=0
k
t 2k−
=...
Inverse triangle matrices, L−1 and U−1
For the lower triangle Li j =
i
j
,
L−1
i j
= (−1)i+j i
j
, (32)
and for the upper ...
Ux = L−1b
L−1
b
i
=
∞
j=0
(−1)i+j i
j
1
2j + 1
=
∞
j=0
(−1)i+j i
j
1
0
x2j
dx
=(−1)i
1
0
1 − x2 i
dx
=(−1)i (2i)!!
(2i + 1...
Integral and double factorial
We apply the integration by parts:
In =
1
0
1 + x2 n
dx
=
1
0
x 1 + x2 n
dx
= x 1 − x2 n 1
0...
The solution, x = U−1L−1b
The last element of unknown vector x(p) is now available in
x(p)
p = L−1
b
p
/ [U]pp =
1
(−2)p
(...
Riordan array
For the proof of our matrix form L−1 and U−1, we make a brief review
of Riordan arrays.
A Riordan array is a...
Matrix multiplication
By definition, the product of two Riordan arrays is
R d(t), h(t) R a(t), b(t)
n,k
=
∞
j=0
[tn
]d(t) (...
Riordan group
Identity:
R 1, t
n,k
= [tn
]tk
= δnk (42)
Inverse:
R d(t), h(t)
−1
= R
1
d(¯h(t))
, ¯h(t) (43)
where ¯h(t) i...
Maxima script
Since I am not a mathematician but rather a programmer as you know,
I need concrete examples and numeric dum...
Pascal matrix (the matrix L) I
We consider d(t) = 1/(1 − t) and h(t) = t/(1 − t),
L = R
1
1 − t
,
t
1 − t
(44)
The element...
Pascal matrix (the matrix L) II
Now we seek for its inverse.
h =
t
1 − t
⇐⇒ t =
h
1 + h
⇐⇒ ¯h(t) =
t
1 + t
(46)
Thus,
L−1
...
Pascal matrix (the matrix L) III
The inverse relation to the infinite triangle matrices is valid for the
finite ones, thus, ...
Catalan’s triangle (the matrix U) I
Let us consider special case that d(t) = 1, and set h(t) = t(1 − t).
B =R 1, t(1 − t) ...
Catalan’s triangle (the matrix U) II
Now, let us find the inverse.
h = t(1 − t) ⇐⇒ t =
1 ±
√
1 − 4h
2
, but ¯h(t) =
1 −
√
1...
Catalan’s triangle (the matrix U) III
Proof. If we put
w(t) =
1 −
√
1 − 4t
2
(53)
Then, t =
w
1 − w
or w =
t
1 − w
. What ...
Catalan’s triangle (the matrix U) IV
[tn
] wk
w =
t
1 − w
=
1
n
[tn−1
]ktk−1
(1 − t)−n
=
k
n
[tn−k
](1 − t)−n
=
k
n
(−1)n−...
Catalan’s triangle (the matrix U) V
Just transposing and multiplying diagonal matrices from our result
({ai bj Mi j }−1 = ...
Lagrange inverse theorem I
From “Handbook of Mathematical Functions”
If y = f (x), y0 = f (x0), f (x0) 0, then
3.6.6
x = x...
Lagrange inverse theorem II
This is the last one piece for which we need to see a proof.
As an inverse of a Riordan array ...
Lagrange inverse theorem III
We start the multiplication
vn,k =
∞
j=0
j
n
[tn−j
]
t
h(t)
n
[yj
] (h(y))k
, (61)
and by app...
Lagrange inverse theorem IV
When k = n,
vn,n =[tn
]tn
t
h (t)
h(t)
=[t0
]
h (t)
h(t)/t
=[t0
]
h1 + 2h2t + 3h3t2 + · · ·
h1...
Lagrange inverse theorem V
Let us set φ(t) = t/h(t) and w = ¯h(t). Then w(t) is an implicit
function of t such that w = tφ...
Summary
Combinatorics, a branch of mathematics, turned out to be a
powerful tool to study numerical integrators for the N-...
Upcoming SlideShare
Loading in …5
×

Hermite integrators and Riordan arrays

697 views

Published on

2016 Jan. 6, lab seminar

(Jan. 6: corrected typo)

Published in: Software
  • This is an excellent illustration of using the method of coefficients and Riordan arrays. I think it's about time for some free software to support these manipulations. I posted a request for CoQ package that would validate "Method of Coefficients" reasoning but received no feedback. I guess I will try myself; someday. This is a logical basis for supporting automated dealings with Riordan arrays.
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here

Hermite integrators and Riordan arrays

  1. 1. Hermite integrators and Riordan arrays When one feels tired in round-off error Keigo Nitadori keigo@riken.jp Co-Design Team, Flagship 2020 Project RIKEN Advanced Institute for Computational Science January 7, 2016
  2. 2. Today I talk about A general form of the correctors for the family of 2-step Hermite integrators Up to p-th order derivative of the force is calculated directly to obtain 2(p + 1)-th order accuracy A mathematical proof in elementary algebra is presented In my last talk in MODEST 15-S, a simple and beautiful proof made by Satoko Yamamoto were mostly omitted, but today I would like you to see the full story.
  3. 3. References I appreciate for many useful online documents: 1. R. Sprugnoli (2006). An Introduction to Mathematical Methods in Combinatorics. http://www.dsi.unifi.it/~resp/Handbook.pdf 2. R. Sprugnoli (2006). Riordan Array Proofs of Identities in Gould’s Book. http://www.dsi.unifi.it/~resp/GouldBK.pdf 3. D. Merlini (2011). A survey on Riordan arrays. https://lipn.univ-paris13.fr/~banderier/ Seminaires/Slides/merlini.pdf 4. P. Barry (2005) A Catalan Transform and Related Transformations on Integer Sequences. Journal of Integer Sequences 8. https://cs.uwaterloo.ca/journals/JIS/ VOL8/Barry/barry84.html
  4. 4. Polynomial shift by Pascal matrix I We consider to shift a finite order polynomial f (t), of which up to the p-th derivatives are f (n) (t) = dn dtn f (t) (0 ≤ n ≤ p). Adjusting the dimension of the derivatives by a step size h, a vector F(t) = f (t) h f (1) (t) h2 f (2) (t)/2! ... hn f (p) (t)/n! . (1) obeys a differential equation d dt F(t) = 1 h 0 1 0 · · · 0 0 0 2 · · · 0 0 0 0 ... ... ... ... ... ... p 0 0 0 · · · 0 F(t). (2)
  5. 5. Polynomial shift by Pascal matrix II A formal solution at t + h is, F(t + h) = exp  0 1 0 · · · 0 0 0 2 · · · 0 0 0 0 ... ... ... ... ... ... p 0 0 0 · · · 0  F(t) = 0 0 1 0 2 0 · · · p 0 0 1 1 2 1 · · · p 1 0 0 2 2 · · · p 2 ... ... ... ... ... 0 0 0 · · · p p F(t). (3) This matrix is referred to as upper triangle Pascal matrix.
  6. 6. Polynomial shift by Pascal matrix III Example For p = 9, 1 1 1 1 1 1 1 1 1 1 0 1 2 3 4 5 6 7 8 9 0 0 1 3 6 10 15 21 28 36 0 0 0 1 4 10 20 35 56 84 0 0 0 0 1 5 15 35 70 126 0 0 0 0 0 1 6 21 56 126 0 0 0 0 0 0 1 7 28 84 0 0 0 0 0 0 0 1 8 36 0 0 0 0 0 0 0 0 1 9 0 0 0 0 0 0 0 0 0 1
  7. 7. Construction of the Hermite integrators I For simplicity, we integrate from t = −h to t = h, thus a timestep becomes ∆t = 2h (not h). The inputs are additions and subtractions F+ n = 1 2 hn n! f (n) (h) + f (n) (−h) , (4) F− n = 1 2 hn n! f (n) (h) − f (n) (−h) , (5) for 0 ≤ n ≤ p. The outputs are fitting polynomial at t = 0, Fn = hn n! f (n) (0), (6) for 0 ≤ n ≤ 2p + 1. The integral includes only the even order terms: ∆v = h −h f (t)dt = F0 + 1 3 F2 + 1 5 F4 + · · · + 1 2p + 1 F2p, (7) for a 2(p + 1)-th order integrator.
  8. 8. Construction of the Hermite integrators II A linear equation we want to solve is F+ 0 F− 1 F+ 2 F− 3 ... = 0 0 2 0 4 0 6 0 . . . 0 2 1 4 1 6 1 . . . 0 2 2 4 2 6 2 . . . 0 0 4 3 6 3 . . . ... ... ... ... ... F0 F2 F4 F6 ... , (8) of size (p + 1). The element of matrix is 2j i (counted from (0, 0)), which means we have picked up the even number columns from the Pascal matrix.
  9. 9. Construction of the Hermite integrators III Example By solving F+ 0 F− 1 F+ 2 = 1 1 1 0 2 4 0 1 6 F0 F2 F4 , (9) we have F0 F2 F4 = 1 8 8 −5 2 0 6 −4 0 −1 2 F+ 0 F− 1 F+ 2 . (10) The sixth-order integrator is thus ∆v =2 F0 + 1 3 F2 + 1 5 F4 h = F+ 0 − 2 5 F− 1 + 2 15 F+ 2 ∆t (11)
  10. 10. Coefficients table F+ 0 F− 1 F+ 2 F− 3 F+ 4 F− 5 F+ 6 F− 7 A2 1 H4 1 − 1 3 H6 1 − 2 5 2 15 H8 1 − 3 7 4 21 − 2 35 H10 1 − 4 9 6 27 − 2 21 8 315 H12 1 − 5 11 8 33 − 4 33 8 165 − 8 693 H14 1 − 6 13 10 39 − 20 143 48 715 − 32 1287 16 3003 H16 1 − 7 15 12 45 − 2 13 16 195 − 16 429 64 5005 − 16 6435 Table: Coefficients for up to the 16th-order Hermite integrator
  11. 11. General form For the 2(p + 1)-th order integrator (p ≥ 0), the k-th term (0 ≤ k ≤ p) is, c(p) k = 1 (−2)k (2k)!! (2k + 1)!! p − k + m p − k (2k + 1)!! (2k + 1 − 2m)!! (2p + 1 − 2m)!! (2p + 1)!! , (12) with m = (k + 1)/2 . A recurrent form starting from the diagonal element to the bottom is c(p) k =    1 (−2)p (2p)!! (2p + 1)!! (p = k) p − k + m p − k 2p + 1 − 2m 2p + 1 c(p−1) k (p > k) . (13) It is a rational recurrent form, however, we still seek for a differential recurrent form in c(p) k − c(p−1) k for our proof.
  12. 12. Coffee break (double factorial) Definition: n!! = n · (n − 2)!! (for n ≥ 2), 1!! = 0!! = 1. (14) (There are products of odd numbers, and even numbers) Properties: (2n)!! =2n n!, (15) (2n + 1)!!(2n)!! =(2n + 1)!, (16) (2n + 1)!! = (2n + 1)! 2nn! = (n + 1)! 2n 2n + 1 n , (17) (2n)!!(2n − 1)!! =(2n)!, (18) (2n − 1)!! = (2n)! 2nn! = n! 2n 2n n = n! 2n 2n n 2n − 1 n − 1 , (19) etc.
  13. 13. Outline of the proof 1. We wrote an expected form of the coefficients, c(p) k 2. Then calculate a differential recurrence, c(p) k − c(p−1) k 3. We set a linear equation Ax = b, of which solution x should correspond to the expected coefficients 4. LU decomposition, A = LU 5. The proof for the form of inverse matrices L−1 and U−1 will be shown later, by using of a modern tool Riordan arrays 6. Calculate L−1b 7. Finally we see that the solution x = U−1L−1b obeys the same recursion to c(p) k , given a matrix size (p + 1)
  14. 14. Differential recurrence I This is a hand exercise (remember m = (k + 1)/2 ): c(p) k − c(p−1) k =c(p) k 1 − p − k + m p − k 2p + 1 − 2m 2p + 1 =c(p) k m(2k + 1 − 2m) (p − k + m)(2p + 1 − 2m) (20) Now, the difference is simplified as in 1 (−2)k (2k)!! (2k + 1)!! p − k + m m (2k + 1)!! (2k + 1 − 2m)!! (2p + 1 − 2m)!! (2p + 1)!! × m(2k + 1 − 2m) (p − k + m)(2p + 1 − 2m) = (−1)k k! (2p + 1)!! p − k + m − 1 m − 1 (2p − 1 − 2m)!! (2k − 1 − 2m)!! = (−1)k k! (2p + 1)!! (p − k + m − 1)! (p − k)!(m − 1)! 2k−m 2p−m (k − m)! (p − m)! (2p − 2m)! (2k − 2m)! (21)
  15. 15. Differential recurrence II Let us now introduce b = k mod 2 ∈ {0, 1}, and ¯b = 1 − b, hence k = 2m − b. Some useful properties are: (n + b)! n! = (n + 1)b , (n − b)! (n − 1)! = n ¯b , (n − 1)b a ¯b = n − b, (−1)k 2p−k (2m − b)! (2p + 1)!! (p − m − 1 + b)! (p − 2m + b)!(m − 1)! (m − b)! (p − m)! (2p − 2m)! (2m − 2b)! = (−1)k 2p−k (2m − 1)b (2p + 1)!! (2p − 2m)! (p − 2m + b)! m ¯b (p − m)¯b × 2 ¯b 2¯b = (−1)k 2p−k (2p + 1)!! (2m − 1)b (2m) ¯b (p − 2m + b)! (2p − 2m)! (2p − 2m)¯b = (−1)k 2p−k (2p + 1)!! 2m − b (p − 2m + b)! (2p − 2m + b − 1)! × (2p)!! 2p p(p − 1)! = (−1)k 22p−k (2p)!! (2p + 1)!! k p (2p − k − 1)! (p − k)!(p − 1)! (22)
  16. 16. Differential recurrence III Finally, we have c(p) k =    1 (−2)p (2p)!! (2p + 1)!! (p = k) c(p−1) k + (−1)k 22p−k (2p)!! (2p + 1)!! k p 2p − k − 1 p − k (p > k) . (23) for which we are going to make the proof. Note that this form does not include m = (k + 1)/2 .
  17. 17. Differential recurrence (old version) I When k is even (k = 2m), (2m)! (2p + 1)!! p − m − 1 m − 1 (2p − 2m − 1)!! (2m − 1)!! = (2m)!! (2p + 1)!! (p − m − 1)! (p − 2m)!(m − 1)! (2p − 2m − 1)! (2p − 2m − 2)!! = 1 2p−m−1 2mm! (2p + 1)!! (2p − 2m − 1)! (p − 2m)!(m − 1)! × m! m! = k 2p−k 1 (2p + 1)!! (2p − 2m − 1)! (p − 2m)! × (p − 1)! (p − 1)! = k 2p−k (p − 1)! (2p + 1)!! 2p − k − 1 p − k × p p (2p)!! (2p)!! = k 2p−k (2p)!! (2p + 1)!! (p − 1)! 2p p! 2p − k − 1 p − k = 1 22p−k (2p)!! (2p + 1)!! k p 2p − k − 1 p − k . (24)
  18. 18. Differential recurrence (old version) II When k is odd (k = 2m − 1), − (2m − 1)! (2p + 1)!! p − m m − 1 (2p − 2m − 1)!! (2m − 3)!! = − (2m − 1)(2m − 2)!! (2p + 1)!! (p − m)(p − m − 1)! (p − 2m + 1)!(m − 1)! (2p − 2m − 1)! (2p − 2m − 2)!! = − 2m−1 2p−m−1 (2m − 1)(p − m) (2p + 1)!! (2p − 2m − 1)! (p − 2m + 1)! × 2p − 2m 2p − 2m = − 1 2p−k−1 k (2p + 1)!! p − m 2p − 2m (2p − 2m)! (p − 2m + 1)! × (p − 1)! (p − 1)! = − k 2p−k (p − 1)! (2p + 1)!! 2p − k − 1 p − k × p p (2p)!! (2p)!! = − 1 22p−k (2p)!! (2p + 1)!! k p 2p − k − 1 p − k . (25)
  19. 19. Linear equation to solve What we want to solve is Ax = b, (A ∈ N (p+1)×(p+1) 0 , x, b ∈ Qp+1 ) (26) where Ai j = 2i j and bj = 1 2j + 1 , (0 ≤ i, j ≤ p) (27) For example when p = 5, 1 0 0 0 0 0 1 2 1 0 0 0 1 4 6 4 1 0 1 6 15 20 15 6 1 8 28 56 70 56 1 10 45 120 210 252 1 −5/11 8/33 −4/33 8/165 −8/693 = 1 1/3 1/5 1/7 1/9 1/11 . (28) gives the coefficients of the 12th-order integrator
  20. 20. LU decomposition An LU decomposition A = LU is available in, Ai j = 2i j , Li j = i j , Ui j = 22i−j i j − i , (29) irrespective to the matrix size (p + 1). Thus, LUx = b can be solved as x = U−1L−1b. Example 1 0 0 0 0 0 1 2 1 0 0 0 1 4 6 4 1 0 1 6 15 20 15 6 1 8 28 56 70 56 1 10 45 120 210 252 = 1 0 0 0 0 0 1 1 0 0 0 0 1 2 1 0 0 0 1 3 3 1 0 0 1 4 6 4 1 0 1 5 10 10 5 1 1 0 0 0 0 0 0 2 1 0 0 0 0 0 4 4 1 0 0 0 0 8 12 6 0 0 0 0 16 32 0 0 0 0 0 32 (30) For the proof, we prepare the tools in mathematical combinatorics.
  21. 21. Formal power series, and coefficient extraction Formal power series f (t) = ∞ k=0 tk fk f (t) is referred to as a generating function of a sequence ( f0, f1, f2, . . .). Coefficient extraction [tn ] f (t) = fn This is an operator but with a weak associativity. Shifting [tn ]tk f (t) = [tn−k ] f (t) Newton’s binomial theorem [tk ](1 + t)n = n k
  22. 22. Proof for A = LU 2i j =[tj ](1 + t)2i =[tj ] 1 + t(t + 2) i =[tj ] ∞ k=0 i k tk (t + 2)k =[tj−k ] ∞ k=0 i k ∞ =0 k t 2k− = ∞ k=0 i k k j − k 22k−j . (31) This discussion is valid for finite matrices, for k iterates from 0 to min(i, j), touching only the non-zero elements of the upper and lower triangular matrices.
  23. 23. Inverse triangle matrices, L−1 and U−1 For the lower triangle Li j = i j , L−1 i j = (−1)i+j i j , (32) and for the upper triangle Ui j = 22i−j i j − i , [U−1 ]i j =    (−1)i+j 22j−i i j 2j − i − 1 j − i (1 ≤ i ≤ j) 1 (i = j = 0) 0 (otherwise) . (33) The both do not depend on the matrix size 0 ≤ i, j ≤ p. The proof is a little bit technical and we put them later.
  24. 24. Ux = L−1b L−1 b i = ∞ j=0 (−1)i+j i j 1 2j + 1 = ∞ j=0 (−1)i+j i j 1 0 x2j dx =(−1)i 1 0 1 − x2 i dx =(−1)i (2i)!! (2i + 1)!! . (34) This vector does not depend on the vector length p. A proof of the fourth identity (an integral to double factorials) is in the next slide.
  25. 25. Integral and double factorial We apply the integration by parts: In = 1 0 1 + x2 n dx = 1 0 x 1 + x2 n dx = x 1 − x2 n 1 0 − 1 0 x 1 − x2 n dx = − 2n 1 0 −x2 1 + x2 n−1 dx = − 2n 1 0 1 − x2 1 + x2 n−1 − 1 + x2 n−1 dx = − 2n (In − In−1) . (35) For In = 2n 2n + 1 In−1 and I0 = 1, we have In = (2n)!! (2n + 1)!! .
  26. 26. The solution, x = U−1L−1b The last element of unknown vector x(p) is now available in x(p) p = L−1 b p / [U]pp = 1 (−2)p (2p)!! (2p + 1)!! . (36) In a general case, x(p) k = p j=k U−1 k j L−1 b j . (37) From the p-dependency of the solution vector x(p), we have a recursion x(p) k − x(p−1) k = U−1 k p L−1 b p = (−1)k 22p−k k p (2p)!! (2p + 1)!! 2p − k − 1 p − k . (38) This is (23), exactly what we wanted!!
  27. 27. Riordan array For the proof of our matrix form L−1 and U−1, we make a brief review of Riordan arrays. A Riordan array is an infinite lower triangular matrix taking a pair of formal power series d(t) (d0 0) and h(t) (h0 = 0, h1 0) 1, of which element in the n-th row k-th column (counted from (0, 0)) is R d(t), h(t) n,k = [tn ]d(t) (h(t))k (39) They make a group, providing explicit forms of matrix multiplication, identity, and inverse. In this study, we exploit the Riordan arrays to find matrix inverses of which coefficients are expressed in binomials. 1Notice that Handbook.pdf by Sprugnoli takes a slightly different notation R(d(t), th(t)) with d0, h0 0, which does not make a practical difference.
  28. 28. Matrix multiplication By definition, the product of two Riordan arrays is R d(t), h(t) R a(t), b(t) n,k = ∞ j=0 [tn ]d(t) (h(t))j · [yj ]a(y) (b(y))k = [tn ]d(t) ∞ j=0 (h(t))j · [yj ]a(y) (b(y))k = [tn ]d(t) · a (h(t)) (b (h(t)))k = R d(t) · a(h(t)), b(h(t)) n,k . (40) In the third equality, we used a relation ∞ j=0 xj · [yj ] f (y) = f (x), (41) with x = (h(t))j and f (y) = a(y)(b(y))k .
  29. 29. Riordan group Identity: R 1, t n,k = [tn ]tk = δnk (42) Inverse: R d(t), h(t) −1 = R 1 d(¯h(t)) , ¯h(t) (43) where ¯h(t) is an inverse function of h(t) such that ¯h(h(t)) = h(¯h(t)) = t. Proof. Just substitute a(t) = 1/d(¯h(t)) and b(t) = ¯h(t) for the matrix multiplication to obtain an identity R(1, t) as a product.
  30. 30. Maxima script Since I am not a mathematician but rather a programmer as you know, I need concrete examples and numeric dumps to study the arrays. Maxima is a nice free software for such purposes. RAelem(d, h, n, k) := coeff(taylor(d*h^k, t, 0, n), t^n); genRA(d, h, nmax) := genmatrix( lambda([i,j], RAelem(d, h, i-1, j-1)), nmax, nmax); Today, these software are not clever enough to find a general form or a proof automatically, but they help us greatly.
  31. 31. Pascal matrix (the matrix L) I We consider d(t) = 1/(1 − t) and h(t) = t/(1 − t), L = R 1 1 − t , t 1 − t (44) The elements are binomial coefficients: Ln,k = [tn ] 1 1 − t t 1 − t k =[tn−k ](1 − t)−k−1 =[tn−k ] ∞ =0 −k − 1 1−k−1− (−t) =(−1)n−k −k − 1 n − k = n k . (45) Here, we used −n k = (−1)k n+k−1 k .
  32. 32. Pascal matrix (the matrix L) II Now we seek for its inverse. h = t 1 − t ⇐⇒ t = h 1 + h ⇐⇒ ¯h(t) = t 1 + t (46) Thus, L−1 = R 1 1 + t , t 1 + t (47) L−1 n,k [tn ] 1 1 + t t 1 + t k =[tn−k ](1 + t)−k−1 = −k − 1 n − k =(−1)n−k n k (48)
  33. 33. Pascal matrix (the matrix L) III The inverse relation to the infinite triangle matrices is valid for the finite ones, thus, for example: 1 0 0 0 0 0 1 1 0 0 0 0 1 2 1 0 0 0 1 3 3 1 0 0 1 4 6 4 1 0 1 5 10 10 5 1 −1 = 1 0 0 0 0 0 −1 1 0 0 0 0 1 −2 1 0 0 0 −1 3 −3 1 0 0 1 −4 6 −4 1 0 −1 5 −10 10 −5 1 This confirms the relation Li j = i j and [L−1]i j = (−1)i+j i j in the main story.
  34. 34. Catalan’s triangle (the matrix U) I Let us consider special case that d(t) = 1, and set h(t) = t(1 − t). B =R 1, t(1 − t) (49) Bn,k =[tn ](t(1 − t))k =[tn−k ](1 − t)k =(−1)n−k k n − k (50) 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 −2 1 0 0 0 0 0 0 1 −3 1 0 0 0 0 0 0 3 −4 1 0 0 0 0 0 −1 6 −5 1 0 0 0 0 0 −4 10 −6 1
  35. 35. Catalan’s triangle (the matrix U) II Now, let us find the inverse. h = t(1 − t) ⇐⇒ t = 1 ± √ 1 − 4h 2 , but ¯h(t) = 1 − √ 1 − 4t 2 B−1 =R 1, 1 − √ 1 − 4t 2 (51) B−1 n,k = [tn ] 1 − √ 1 − 4t 2 k = k n 2n − k − 1 n − k (52) for 1 ≤ k ≤ n, otherwise δnk . 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 2 2 1 0 0 0 0 0 5 5 3 1 0 0 0 0 14 14 9 4 1 0 0 0 42 42 28 14 5 1 0 0 132 132 90 48 20 6 1
  36. 36. Catalan’s triangle (the matrix U) III Proof. If we put w(t) = 1 − √ 1 − 4t 2 (53) Then, t = w 1 − w or w = t 1 − w . What we want is [B−1]n,k = [tn](w(t))k and we can apply the following Lagrange inverse theorem: [tn ] F(w) w = tφ(w) = 1 n [tn−1 ]F (t) φ(t) n (54) In our case, F(w) = wk and φ(t) = (1 − t)−1.
  37. 37. Catalan’s triangle (the matrix U) IV [tn ] wk w = t 1 − w = 1 n [tn−1 ]ktk−1 (1 − t)−n = k n [tn−k ](1 − t)−n = k n (−1)n−k −n n − k = k n 2n − k − 1 n − k (55) for 1 ≤ k ≤ n. For other cases n = 0, k = 0, or k > n, we have trivial elements [B−1]n,k = δnk .
  38. 38. Catalan’s triangle (the matrix U) V Just transposing and multiplying diagonal matrices from our result ({ai bj Mi j }−1 = {a−j b−i[M−1]i j }) Bi j =(−1)i+j j i − j , B−1 i j = j i 2i − j − 1 i − j for 1 ≤ j ≤ i otherwise δi j, we have the last piece needed in our study: Ui j =(−1)i+j 22i−j Bji = 22i−j i j − i [U−1 ]i j = (−1)i+j 22j−i B−1 ji = (−1)i+j 22j−i i j 2j − i − 1 j − i for 1 ≤ i ≤ j otherwise δi j, except that Lagrange inverse theorem is used without proof.
  39. 39. Lagrange inverse theorem I From “Handbook of Mathematical Functions” If y = f (x), y0 = f (x0), f (x0) 0, then 3.6.6 x = x0 + ∞ k=1 (y − y0)k k!  dk−1 dxk−1 x − x0 f (x) − y0 k x=x0 3.6.7 g(x) = g(x0) + ∞ k=1 (y − y0)k k!  dk−1 dxk−1 g (x) x − x0 f (x) − y0 k x=x0 where g(x) is any function indefinitely differentiable. In formal power series, [tn ] w(t) w = tφ(w) = 1 n [tn−1 ] φ(t) n , (56) [tn ] F(w) w = tφ(w) = 1 n [tn−1 ]F (t) φ(t) n . (57)
  40. 40. Lagrange inverse theorem II This is the last one piece for which we need to see a proof. As an inverse of a Riordan array R(1, h(t)), that is R(1, ¯h(t)) for d(t) = 1, we assume the following form: dn,k = R 1, ¯h(t) n,k = k n [tn−k ] t h(t) n . (58) If the following relation vn,k = ∞ j=0 dn, j[yj ] (h(y))k = δnk, (59) if satisfied, it is equivalent to [tn ] ¯h(t) k = R (1, h(t))−1 n,k = R 1, ¯h(t) n,k = k n [tn−k ] t h(t) n . (60)
  41. 41. Lagrange inverse theorem III We start the multiplication vn,k = ∞ j=0 j n [tn−j ] t h(t) n [yj ] (h(y))k , (61) and by applying a differentiation rule j[yj ] (h(y))k = [yj−1 ] (h(y))k = [yj ]y h (y) k (h(y))k−1 , (62) we continue the calculation, as in vn,k = k n ∞ j=0 [tn−j ] t h(t) n [yj ]yh (y) (h(y))k−1 = k n [tn ] t h(t) n th (t) (h(t))k−1 . (63) Here, we used a convolution rule: [tn ] f (t)g(t) = ∞ j=0 [tj ] f (t) · [yn−j ]g(y). (64)
  42. 42. Lagrange inverse theorem IV When k = n, vn,n =[tn ]tn t h (t) h(t) =[t0 ] h (t) h(t)/t =[t0 ] h1 + 2h2t + 3h3t2 + · · · h1 + h2t + h3t2 + · · · = 1 (65) When k n, vn,k = k n [tn ]tn+1 (h(t))k−n−1 h (t) = k n [t−1 ] 1 k − n (h(t))k−n = 0 (66) Here, (h(t))k−n is a formal Laurent series of which derivative cannot contain a non-zero coefficient for t−1.
  43. 43. Lagrange inverse theorem V Let us set φ(t) = t/h(t) and w = ¯h(t). Then w(t) is an implicit function of t such that w = tφ(w). For a special case k = 1, we have [tn ] w(t) w = tφ(w) = dn,1 = 1 n [tn−1 ] φ(t) n . In more general cases, for a formal power series F(w), we have [tn ] F(w) w = tφ(w) =[tn ] ∞ k=0 Fk wk = ∞ k=0 Fk dn,k = ∞ k=0 Fk k n [tn−k ] φ(t) n = 1 n [tn−1 ] ∞ k=0 kFktk−1 φ(t) n = 1 n [tn−1 ]F (t) φ(t) n .
  44. 44. Summary Combinatorics, a branch of mathematics, turned out to be a powerful tool to study numerical integrators for the N-body problem. Especially the formal power series, Lagrange inverse theorem and modern Riordan arrays are practical tools for even non-mathematicians. LU decomposition is not only for Top500/Green500, but a general tool for studying integers. Mathematics is useful. Satoko Yamamoto did a great work.

×